Birne Binegar: I guess we'll start today what we're going to do is we're going to continue on with our discussion of linear transformation.
Birne Binegar: What we saw on went on Monday was that the linear transformation they're related to matrices of review that the start.
Birne Binegar: Then, just like matrices we could associate sub spaces to matrices similarly will be able to assign.
Birne Binegar: To sub spaces to a linear transformation and, in fact, because of the connection with with matrices will actually be able to find bases for the sub spaces.
Birne Binegar: And if I have time i'll talk about the dimension of these two sub spaces and how you actually combine multiple linear transformations to create to create other linear transformations which kind of mathematical but actually part of the point.
Birne Binegar: Of linear algebra and predict particular but today's lecture is that relate these abstract things probably your confirmation to concrete calculations setting that's why we bring in matrices because they give us numbers that we can actually work with rather than abstract ideas.
Birne Binegar: So let me, let me start.
Birne Binegar: Our remind you that a linear transformation is a function between two vectors spaces and it's not an arbitrary function will require it to be compatible with our vector space operations.
Birne Binegar: So, in particular, we wanted to behave nicely with respect scale multiplication of vectors and so here if we multiply a vector before we apply the function.
Birne Binegar: We get the same result it if we had applied the function to the bear vector and then do the scale or multiplication so inside the parentheses we're doing.
Birne Binegar: scale multiplication of vectors inside the domain or M and then on the left, right hand side of this equation we're doing the scale or multiplication.
Birne Binegar: After we've already reached n dimensional vectors by the application of that function and, similarly, we want it to behave well respected vector edition so whether we.
Birne Binegar: Add vectors before or after we apply the linear.
Birne Binegar: transformation, we want to get the same result.
Birne Binegar: General functions are not going to have these very special properties and, in fact, these are so special that the functions end up being really simple, basically, the only way, you can create such a function is through them bite bite.
Birne Binegar: Essentially, they have to be functions that are related to matrix multiplication and i'll show you that next, first of all, if you have a.
Birne Binegar: matrix with in rows and columns you can use that matrix to create a function that will take you from an m dimensional vector space to an n dimensional vector space, all you need is matrix multiplication.
Birne Binegar: So how is that going to work.
Birne Binegar: So you take a vector that lives inside on em and for this you create an m by one column vector.
Birne Binegar: I take that back here and i'm wondering interpreted as a matrix so I can perform matrix multiplication, then I multiply.
Birne Binegar: These n dimensional column vectors I multiply them by.
Birne Binegar: This n by n matrix because the number of columns is equal, the number of rows that multiplication is defined and what it will produce.
Birne Binegar: it's another column vector but now, one that has in separate rosen and that's you then interpret as an element of our end.
Birne Binegar: So what this matrix multiplication is doing is taking you from n dimensional vectors and it's.
Birne Binegar: Taking out as a result n dimensional vector so we do get a map between these two vector space is using matrix multiplication.
Birne Binegar: And in fact it's easy to verify that the function that you define this way naturally satisfies these two properties because scale multiplication is going to commute with.
Birne Binegar: matrix multiplication and when you multiply a sum of vectors by a matrix you just hit that matrix multiplication just distributes over that son X plus y is just going to equal a X, plus a why distribute the matrix multiplication distribute over vector edition.
Birne Binegar: So, so these two properties are going to be automatic when I create a function, using a matrix.
Birne Binegar: What we saw last time, is that we can go in the opposite direction as well, if I have a function that satisfies these two properties, I can construct a matrix that will have the same action, the way this guy acts on a vector is going to be the same thing as a matrix multiplication.
Birne Binegar: The only thing I lack on this slide so far is a recipe for how to construct this matrix from a given function satisfying those properties, but here's the recipe what we do is first of all.
Birne Binegar: We take.
Birne Binegar: Anyone em set of em vectors that's going to be the standard.
Birne Binegar: basis for an m dimensional vector space, why we get these factors is we just use a natural coordinate directions to going to equal 010 etc So these are the standard basis vectors taking linear combinations of these factors, we can reach any point insight on em.
Birne Binegar: And so what we do, since easy these guys are easy to create here's the recipe, we can.
Birne Binegar: Each of them is going to be an m m dimensional vector so we can apply the transformation to it that's going to take us over to a vector within entries and we write them vertically.
Birne Binegar: That creates a matrix with and rose for each component, where we land and then we have a centrally and different starting point with we have in this m dimensional space we're going to have exactly am standard basis factors.
Birne Binegar: so soon as you got the function satisfying these properties, you can construct this matrix.
Birne Binegar: And for given function, this is just going to be a matrix of numbers and that's that's what we're trying to do we're trying to take this abstract definition and bring it into a calculation that will set a calculable setting.
Birne Binegar: Alright, so i'm.
Birne Binegar: That much we covered last time and, in fact, with a little bit of what i'm about to say we covered last time as well.
Birne Binegar: Just like we did for.
Birne Binegar: For matrix season.
Birne Binegar: we're going to attach sub spaces to function between to back your spaces that behave like a linear transformation.
Birne Binegar: For matrix where we got sub spaces, is well, we can look at the robe actors of this matrix and take all possible linear combinations of them that will create a subspace of well each row is going to have exactly an entry, so this is going to be a subspace on our LM.
Birne Binegar: Or alternatively, we can focus on the column vectors of the matrix each column, of which there I am it's going to have exactly as many entries in that column, as you have rows.
Birne Binegar: And so the column vector is all going to be n dimensional vectors there span is going to be a subspace of the vector space and dimensional vectors.
Birne Binegar: And it will, and finally, we also have another matrix another subspace we can attach to a matrix.
Birne Binegar: Given the matrix we can interpret it as a coefficient matrix of a homogeneous linear system, and then we can look at the solution set up the system that's going to be a subspace of the space of variables.
Birne Binegar: which will be a subspace of our em all right now, last week we talked about each of the sub spaces and how to find bases for these we're going to use that material, because what we've just seen is.
Birne Binegar: We do a given function between two vector space is a linear transformation, we can attach a matrix and so.
Birne Binegar: what's going to turn out to be.
Birne Binegar: matrices the subs basis we attached to a.
Birne Binegar: linear transformation it's actually going to correspond to co op column spaces and no spaces, about the associated matrix so that's what's going to be shown today.
Birne Binegar: But first thing i'm gonna do.
Birne Binegar: Is i'm going to be fine sub spaces attached to a function between two vector space it i'm going to define it in such a way that does not refer to an underlying matrix but rather than only using sort of function theory.
Birne Binegar: So let's let's put the function theory up.
Birne Binegar: Here we have sort of abstract.
Birne Binegar: function theory.
Birne Binegar: Not abstract, in the sense of difficult, but in the sense that it's it's really kind of loosely.
Birne Binegar: defined, we have a set a set be function that takes you from.
Birne Binegar: The set a to the set be, meaning a rule or once you choose a point A inside here you have a rule that specifies a corresponding image point image of a by the function F.
Birne Binegar: Now, when you collect together all the points in the, this is still the Co domain.
Birne Binegar: And the set a is called the domain.
Birne Binegar: Once you have this rule, you can look in the co domain, the target space, so to speak, and you can figure out.
Birne Binegar: All the points in the co domain that are reachable by applying that function it's not necessarily the whole space, if you take a function say F of X let's write it.
Birne Binegar: From the real line to the real line that sent X two X squared well.
Birne Binegar: The image of this function, the set of points in the on the real line that are reachable by this function it's not the entire real line you're only going to reach the positive or actually the non negative.
Birne Binegar: Real numbers, when you square them so whether or not X is positive and negative it square is going to be non negative, so I don't get the entire co domain by an application of this function that's.
Birne Binegar: The image of a function.
Birne Binegar: is precisely the points that you can reach in the co domain by by application of that function so that's going to be the set point B and the co domain, such that B is equal to F of a for some.
Birne Binegar: A inside the domain all right now, this is just.
Birne Binegar: let's not call it abstract let's just call it general function theory, you have a function, you have a Co domain and then you have this image set.
Birne Binegar: What are we going what we're going to look at now is this situation but specialized to the case when our function is between two vectors spaces and the function is a linear transformation.
Birne Binegar: So the range.
Birne Binegar: Of a linear transformation from our m to our end is going to be exactly the same thing.
Birne Binegar: As the image of that.
Birne Binegar: function.
Birne Binegar: But we will set it up it's all this now describe it linear algebraic like it's going to be all the vectors that live sorry it's going to be all the vectors that live in the co domain, so why.
Birne Binegar: inside.
Birne Binegar: Or, in just like we have D inside little be inside be such that y is equal to X.
Birne Binegar: For some.
Birne Binegar: X inside the domain, which is our m.
Birne Binegar: Alright, so that's a set every point of which is an element of our in.
Birne Binegar: What I want to show you is that, in fact.
Birne Binegar: This set.
Birne Binegar: range of a linear transformation is not only a set a subset of rm it's actually going to be a subspace of rem and then all then.
Birne Binegar: Once I show you that i'll show you how we can actually calculate it by constructing the matrix corresponding to linear transformation and looking at the column space of that matrix.
Birne Binegar: Alright, so let's um let's move ahead here a little bit here's the image as I defined it here.
Birne Binegar: And then the range is just a specialization of the notion of images to our linear algebraic situation where a function is going between two vector spaces and it's also a linear transformation.
Birne Binegar: stuff first thing i'll show you is that the subset.
Birne Binegar: notice that every point in it is an element of rm so this range it's going to be a subset of a subset of this vector space, what we want to show it's it's actually a sub space, so we want to check that it's closed under scale or multiplication and closed under vector addition.
Birne Binegar: don't more precisely what we want to show is that if we start off with a number and a vector that lives inside that set.
Birne Binegar: After we do the scale or multiplication we don't leave the seth and, similarly, if we have two vectors that start off in that set we add them together we don't leave that set that's, these are the conditions that make a set us or subset into a sub space.
Birne Binegar: And it's it's kind of analogous to what we did when we define linear transformation constantly linear algebra we're demanding.
Birne Binegar: Some sort of compatibility property between the object that we're trying to describe whether it's subspace or linear transformation and are two fundamental vector space operations scale multiplication and vector addition.
Birne Binegar: Alright, so what i'm going to show you now is.
Birne Binegar: Sets that I construct in this fashion, my looking at all the points I can reach by applying a linear transformation that.
Birne Binegar: Those sets are closed under scale or multiplication and vector addition.
Birne Binegar: So first i'll show you that it's closed on the scale of multiplication so let's start with a point.
Birne Binegar: That lives in this set that means if it actually must have come from some point living inside the domain or m appear place where we start off.
Birne Binegar: And so, if I have a point that lives in the range, it must have come from some exit lives inside our m what i'll do is i'll take this equation.
Birne Binegar: and multiply it by some number Lambda and get directly this, what I want you to observe that, though, because T is a linear transformation.
Birne Binegar: And we have that compatibility with scale or multiplication and the action of the function, I can take this factor of Lambda.
Birne Binegar: And I can move it inside without changing the value so Lambda t of Lambda X is always equal to Lambda times to have X 20 is a linear transformation.
Birne Binegar: But then the extreme side of this equation shows that this new point the scale or multiple of my given point my Lambda is also in the image it's also in the range of tea, because this guy here has been displayed as the value of the function T at a particular point.
Birne Binegar: So that the extreme side of this equation show that Lambda Why is also living in the range because it's a value it's the value of Lambda X.
Birne Binegar: And so, if you have a point in the range and you have an arbitrary number, the scale or multiple of that vector by Lambda still remains inside that set, and so the range of tea is seem to be closed under scale or multiplication.
Birne Binegar: Similarly, we can look at vector edition start off with two points that are.
Birne Binegar: Essentially, come from our domain by the application of this function T so we'll set it up this way well suppose why one came from some guy X X one living inside the domain why to came from X two coming in the domain.
Birne Binegar: This is, these conditions are just stating explicitly what we mean by wanting lie to living inside the range what i'm going to do is i'll add those two equations together and then.
Birne Binegar: that's what we'll get right away that's because, to a linear transformation.
Birne Binegar: This vector addition can either be calculated before I applied the function as we would hear or after I apply the function like it should get the same result if T is a linear transformation, but then that shows you that the some of these two vectors.
Birne Binegar: is still and lives in the image of the function T because.
Birne Binegar: here's the guy the vector X one X two which produces this guy after we apply the function.
Birne Binegar: Right and so now we've seen that, if we start off with two vectors in the range we if we add them together, it still lives inside the range and so the range of closed under vector edition.
Birne Binegar: And now we've got it so subspace because anytime you have a subset of a vector space that's closed under scale and multiplication close under vector addition meeting when you apply those operations you don't leave this set well, that means it's a sub stakes.
Birne Binegar: Alright, so.
Birne Binegar: i've got a SEC or subset of the domain up the range sorry a subset of the co domain that now been demonstrated to be a subspace of the co domain what we'll do next, you will find a basis for it.
Birne Binegar: Alright, so well i'm going to show you next is that you have a linear transformation you create the corresponding matrix as we did on Monday select your by.
Birne Binegar: noticing how tx on the standard basis factors and using those results to construct a numerical matrix what what i'll show you next is that the.
Birne Binegar: This subspace of.
Birne Binegar: Of the Co domain is actually exactly the same thing as the column space of this matrix.
Birne Binegar: All right now i'm going to show you this.
Birne Binegar: indirectly.
Birne Binegar: So.
Birne Binegar: The idea of the argument here.
Birne Binegar: it's gonna be analogous to.
Birne Binegar: If I know that I have two numbers X, Y now that don't think of them a vector just think of them as numbers, if I know that X is less than or equal to y and I also know that, why is less than equal to X well these two inequality can only hold when X is equal to y.
Birne Binegar: Right, so this this is usually not the way you show the two things are the same, but if both of these conditions are true, it must be that X is equal to y well the argument we're going to do to show that the such are the same, it will show you that if.
Birne Binegar: A is a subset of B and B is a subset of a that these two conditions are equivalent of saying that the two sets are the same.
Birne Binegar: And so.
Birne Binegar: that's the way.
Birne Binegar: i'm going to show you that.
Birne Binegar: These two sets are the same i'm going to show you that every element that sits in the range is also in the column space of the associated matrix and that.
Birne Binegar: Every element in the column space of the associated matrix is also lies in the range of the of the given transformation so remember the call and space that's what you get by taking linear combinations of the columns of this matrix the range well that's what you get by figuring out.
Birne Binegar: Where a given linear transformation, let me put the linear transformation picture up here now we're going from our m we're calling it T.
Birne Binegar: going from our am.
Birne Binegar: and
Birne Binegar: And we're ending up in a another vector space are in the last picture we were looking at.
Birne Binegar: The range.
Birne Binegar: T is exactly the point.
Birne Binegar: In the code domain.
Birne Binegar: Such that y equals.
Birne Binegar: X for.
Birne Binegar: Some X in the domain R m.
Birne Binegar: As opposed to the column space.
Birne Binegar: At which is.
Birne Binegar: A set of all possible linear combinations say say see one times the first column at.
Birne Binegar: Plus dot dot dot plus see let's see they'll be exactly m columns calm and a subset of tea, and then I let those coefficient.
Birne Binegar: range over the entire we don't want so when you actually write these two sets down they don't look like one another, but my claim is.
Birne Binegar: They are exactly the same set, even though they don't appear to be.
Birne Binegar: And it's it's actually not so hard to show this correspondence, because in fact the call of at the way they're prescribed this by looking how tx on the standard basis factors.
Birne Binegar: dope um i'll show you this i'll show you that if I grab an element from this set it's automatically going to be in this set and if I grab an element of this said it's automatically interpreted well as an element of this set.
Birne Binegar: And so, this sort of logic, for showing that the two sets are equal work.
Birne Binegar: Alright, so let's start shooting an element that lives in the range of tea.
Birne Binegar: So right it if, since it in the range, it has to come from some vector in the.
Birne Binegar: Some X over here.
Birne Binegar: And it's going to end up.
Birne Binegar: As P of X over here.
Birne Binegar: And what i'm going to show you, is it I have.
Birne Binegar: A vector in the co domain that came from the domain by application of the function.
Birne Binegar: drawn staying there that in fact it's also an element of the set.
Birne Binegar: Well, how does that work you're just as an equation now what i'll do is i'll take the spectrum X and i'll expand it in terms of the standard basis.
Birne Binegar: For the domain R m there'll be exactly m standard basis vectors that will coordinate, is this the sector space.
Birne Binegar: Take an arbitrary vector I expanded in terms of the standard basis, then I use the property that T is a linear transformation to take the sum of vectors and replace it by.
Birne Binegar: Well linear transformations they're compatible with vector edition so effectively.
Birne Binegar: The sum of vectors.
Birne Binegar: upon which I apply this function, can the same value would be occurred, if I took the individual vectors applied the function and then summed everything up.
Birne Binegar: Well that's the second property of the linear transformation, the first property is that behaves well respect to scale or multiplication as well.
Birne Binegar: Whether I scale have multiplied before I apply the function or after I apply the function if I have a linear transformation, I get the same result.
Birne Binegar: But now, given the way this matrix constructed he applied to the standard basis factors it's giving you a definition exactly the columns of this matrix.
Birne Binegar: And so what I have is an element in the range now being expressed as a linear combination of the columns of a.
Birne Binegar: Just numbers times these column vectors so that puts why inside the column space.
Birne Binegar: And so i've shown you what are these conclusions one and the range everything it's in the rain it's also going to live inside the columns basically associated matrix.
Birne Binegar: Next i'll show you that if I take it back to that lives in the column space it corresponds to a vector that also lives inside the range.
Birne Binegar: So take a point in the column states well, by definition, that means it has to be a linear combination of coal and afters let's write it out that way.
Birne Binegar: This column that was on the other hand, from the way this matrix watch define each column corresponds to the application of tea to a standard basis factor of the domain.
Birne Binegar: So just replace these calm that you're like the corresponding application of tea to a standard basis after we get that and now we again we use the properties of linear transformation is scalable applications can be.
Birne Binegar: This term here is exactly equal to this scale multiplication after we apply the linear transformation it's the same thing is.
Birne Binegar: He applied to the scale of multiple at the starting vector anyway.
Birne Binegar: and, similarly, because we're adding up scalable linear transformations are compatible with tradition, I can do the inspector sums after or before I applied the linear transformation and I should get the same result these two steps are properties of the linear transformation.
Birne Binegar: But now we see that w can be obtained by applying T to a particular vector well, that means that this w what is inside the range of T the range.
Birne Binegar: it's all the points you can reach by applying this function so evidently w can be reached by applying key to this particular vector.
Birne Binegar: And so we've shown that every element in the range lives in the column space every element column space lives in the range and the two sets are equal.
Birne Binegar: Okay now that's that's really ideal.
Birne Binegar: My the problem i'm.
Birne Binegar: working towards addressing is, I have this abstract subspace how do I find a basis for it, so I can actually talk.
Birne Binegar: about particular vectors that live inside this this range, how do I coordinate ISIS subspace well I need a basis.
Birne Binegar: Because the sub spaces are, in fact, the same vector space, I can calculate a basis for this and then use it as a basis for the range.
Birne Binegar: And since other thing subspace well we get this corollary, a basis for the range of linear transformation can be found by finding a basis for the column space.
Birne Binegar: And that we learned how to do last week.
Birne Binegar: You remind you.
Birne Binegar: To find a basis.
Birne Binegar: For the common space.
Birne Binegar: of any matrix a whether or not it comes from a linear transformation or not, we had the following procedure you reduce.
Birne Binegar: hey to.
Birne Binegar: Usually, we go all the way through, we do show Echelon for but it suffices to go to row Echelon porn and then.
Birne Binegar: You.
Birne Binegar: know.
Birne Binegar: Where the pivots live.
Birne Binegar: In that row Echelon form.
Birne Binegar: And then.
Birne Binegar: The basis factors for the column space or the pain by grabbing the columns.
Birne Binegar: grab the corresponding columns.
Birne Binegar: Not of the row Echelon form but up the original matrix a so you grab the corresponding columns.
Birne Binegar: A let me do this explicitly within the example.
Birne Binegar: here's a function, so you taking a point where coordinates X one X two and it's producing up.
Birne Binegar: After the application of that is producing a three dimensional vector with coordinate X one plus x two X one minus X two X one.
Birne Binegar: So that's a function clearly for every pair of numbers every element of our to we get a three dimensional vector.
Birne Binegar: What I want to do is find a basis for the range of that function that's going to be defined by this, but the way will calculate it is by finding an equivalent matrix and then calculating it's column space using this algorithm.
Birne Binegar: So first step let's get let's get that corresponding matrix.
Birne Binegar: The way it's defined, if you look at where you're starting and then use the standard basis vectors for that vector space.
Birne Binegar: Again, though standard basis vectors let me put them over here.
Birne Binegar: You create them simply by.
Birne Binegar: Writing down vectors have a single non zero component in a particular slot.
Birne Binegar: and
Birne Binegar: If you have vectors inside.
Birne Binegar: Our am that means there's going to be exactly am plots here the last standard basis that you're.
Birne Binegar: going to be that and so as soon as I know what my.
Birne Binegar: What my domain is I can immediately construct these standard basis factors.
Birne Binegar: One but they're still going to be vectors inside the domain of this function T, so I can then evaluate the function key at those standard basis sectors and i'll get you America results, so now the abstraction has disappeared and we're working with us matrix with just numbers so ventures.
Birne Binegar: Alright, so this matrix is going to be the matrix that corresponds to this function.
Birne Binegar: notice that.
Birne Binegar: Even though in specifying this function I sort of wrote the value out as a horizontal vector.
Birne Binegar: When I construct this matrix it's necessary to write the answer vertically so, for example.
Birne Binegar: The first guy he applied to the first standard basis factor well that's going to be one X one is equal to one, and the second component is equal to zero.
Birne Binegar: So when I substitute one for X one zero for X two i'm going to get one plus zero is one one minus zero is one in one.
Birne Binegar: second component now X two is going to equal one and X one is going to equal to zero so i'll get zero plus one is 110 minus one is negative one, and then X one is equal to zero for the second standard basis vectors so I get zero.
Birne Binegar: So, which you just evaluate the the function at the standard basis factors to get the results and you arrange them as columns.
Birne Binegar: All right now.
Birne Binegar: Because.
Birne Binegar: These two sets are identical to one and now they're.
Birne Binegar: What we'll do is we'll find a basis for this space like they're the same back your space so i'll just use this calculation to find a basis for the columns days and i'll use that basis as a basis for the range, that means.
Birne Binegar: Taking my given matrix reducing it to row Echelon form.
Birne Binegar: Then, noting where the pivots live.
Birne Binegar: that's the first step second step is to figure out where the pivots live, and we have a pivot in the first column in the second column to get the basis.
Birne Binegar: For the column space of this matrix I go back to the original matrix and I grabbed the columns of the original matrix that correspond to the columns of the role Echelon forum where the pivot slip well there pivots and both columns so I grabbed the first column in the second column.
Birne Binegar: And I write them down.
Birne Binegar: i'm writing them down as horizontal vectors.
Birne Binegar: Because in.
Birne Binegar: In this setting i'm thinking of this is going from a two dimensional vectors the three dimensional vectors not necessarily as a two by one matrix going to a three by one matrix i'm thinking these inspectors i'm just writing them horizontally so.
Birne Binegar: This last step, where I write that this column this column vector right right it horizontally that's simply because i'm now interpreting it as a vector rather than his column vector.
Birne Binegar: Three by one matrix.
Birne Binegar: Alright, so that's it that's that's what you do to find a basis for the range of tea you construct the corresponding matrix and then you find a basis for the that make the column space of that matrix.
Birne Binegar: Using the standard algorithms.
Birne Binegar: Okay.
Birne Binegar: let's see i'll keep that picture up.
Birne Binegar: And next thing I want to talk about those that as a second sub space.
Birne Binegar: That we can attach to a function between to back your spaces.
Birne Binegar: This one is again going to be defined, using function theory ideas but we'll quickly related to the matrix that corresponds to our given function.
Birne Binegar: So here's the set we're interested in this case it's going to be the set of all vectors in the domain that gets sent to the zero vector.
Birne Binegar: So this range of tea it's a sub space of the co domain what i'm going to describe next is going to be.
Birne Binegar: The kernel of tea, if my function is taking em dimensional vectors to n dimensional vectors, unlike the range which lives over here the Colonel is going to be a subspace of the domain.
Birne Binegar: And in fact what it is it's a set of points inside your domain that gets sent.
Birne Binegar: To the zero vector so the guys here are going to get sent.
Birne Binegar: Not to.
Birne Binegar: The rain, it is going to land in range, but the point is is that they all go to 1.0 vector.
Birne Binegar: And so.
Birne Binegar: that's the idea of the Colonel it's all the things that sit in the domain that somehow get mapped to the zero vector inside the code domain under the application of your function at.
Birne Binegar: All right, so that's the that's the set every element of this is a element of the domain and so it's a subset of the domain.
Birne Binegar: i'll show you really quickly that it's also.
Birne Binegar: A subspace of that domain, but I won't do it by showing that it's closed under scale multiplication and vector addition.
Birne Binegar: Rather i'm going to show that that this set is actually a subspace of the domain by showing how do we relate it to the original matrix or the associated matrix at and here's here's the statement that the current this kernel the set of vectors in the domain.
Birne Binegar: is actually equal to the null space of this matrix.
Birne Binegar: Remember, no space.
Birne Binegar: of any matrix, whether it comes from a linear transformation or not this is defined as the solution set.
Birne Binegar: Of a linear system associated with this matrix solution, set up a time X is equal to zero.
Birne Binegar: All right, so what i'll show you is that every element in this kernel every vector in the domain that gets sent to the zero vector also satisfies the equation at times that vector is equal to zero.
Birne Binegar: And that's all we have to do is write down the result so.
Birne Binegar: We take a take it back to that lives inside the Colonel the necessarily the definition of what the Colonel is g of X has to equal zero.
Birne Binegar: But if I have a linear transformation, I can also construct this associated matrix.
Birne Binegar: The purpose of this matrix is that the action of the function can be replicated by matrix multiplication so as soon as I construct this matrix this identity is true.
Birne Binegar: But now you see that if you have a vector that gets sent to zero it's that it's multiplication by this matrix X is also equal to zero, so it must satisfy this equation.
Birne Binegar: And so the two sets are going to be exactly the same.
Birne Binegar: And so the conclusion, then, is that the no space of.
Birne Binegar: Of the matrix is going to be exactly the same thing if the Colonel until in fact what we can do is, we can calculate this guy by calculating this instead, which is great because.
Birne Binegar: When we're looking at this guy it's just a matrix of numbers well this guy is defined more abstractly via.
Birne Binegar: via this definition.
Birne Binegar: Alright, so let's do an example we actually compute.
Birne Binegar: The kernel of a linear transformation.
Birne Binegar: I have a new linear transformation, this one that's going to take three dimensional vectors the two dimensional vectors.
Birne Binegar: Now what i've proposed to do in this example is find a basis for the kernel of this linear transformation i'm going to find a basis for this set.
Birne Binegar: Alright, now that turned out to be equivalent to no space of the associated matrix, and so what i'm going to have to do.
Birne Binegar: The find a basis for this set well look at the same set but express the different way, as the solution, set up a linear system and i'll just sold this linear system to find the elements that live here, how do we do that well write down the matrix.
Birne Binegar: And so here I done that he applied to the first standard basis vectors going to be one minus zero.
Birne Binegar: And then the second component is going to be zero minus zero or zero so I just substitute in for one.
Birne Binegar: One of the standard basis factors to get each of these columns and then I evaluate the result me write it vertically matrix is easily constructed now what we'll try to do is find the solution set of this matrix times up.
Birne Binegar: I guess a three dimensional vector has to equal zero.
Birne Binegar: So we take the matrix reroll reduce it to.
Birne Binegar: If reduce your relational and form, so I do elementary zero operations to systematically well the pivots are already equal the one I need to clear out this entry above the pivot to get it and reduce cholesterol on form once I reach reduce relational and form, I convert back to equations.
Birne Binegar: I also look at the new social unformed decide which variables i'm going to regard it's for free is free, in which variables are regardless fix.
Birne Binegar: The fix variables come from columns of the rational on form that have pivots and the calm variable to correspond the columns without tibbetts those are your three variables.
Birne Binegar: We take the equations coming from the matrix matrix and roll Echelon reduce cholesterol inform we express the fixed variables in terms of the free variable then we tried to write down the solution vector sorry that's written incorrectly that solution vector.
Birne Binegar: should have been.
Birne Binegar: I have X.
Birne Binegar: without any restrictions on it it's going to be a three dimensional vector but once I put these two conditions on the components i'll get.
Birne Binegar: X three.
Birne Binegar: X one is going to equal X three x two is going to equal X three x three already equals X Ray and then this is X three times one one alright so every solution it's going to be a scale or multiple this factor and so back the spectrum 111 is.
Birne Binegar: A basis for the solution set.
Birne Binegar: So it's the basis factor.
Birne Binegar: For the solution set.
Birne Binegar: But that.
Birne Binegar: Because the solution set at of X times X is equal to zero, is exactly equal to the kernel of the corresponding linear transformation what we get from that vector is a basis factor.
Birne Binegar: For.
Birne Binegar: The karate because we've seen that that kernel of tea is the same thing.
Birne Binegar: As the north face of the associated matrix so I use the matrix to calculate this this basis vector.
Birne Binegar: And so, in the end I think i've got this yeah this is written in correctly, you should be a vector one one and it has the third component one, so we please ignore the title i'll fix that as soon as I get back to my office.
Birne Binegar: All right, and so there we go, we have to sub spaces attached to a linear transformation.
Birne Binegar: see if I can put a full picture up so you can.
Birne Binegar: See how this works, so we have.
Birne Binegar: We have a linear transformation from em dimensional space coming n dimensional space, this is going to correspond to.
Birne Binegar: An end by am matrix.
Birne Binegar: So that's one idea that we've.
Birne Binegar: been using today, and the second thing is that we have sub spaces, a strong picture let's say we have.
Birne Binegar: Our em now i'm not gonna write it as a circle, let me write some down some coordinate grid, so you think of this more as a as a vector space is being mapped over to a different vector space or in.
Birne Binegar: By the application of some tea that's going from em dimensional vectors n dimensional vectors now we've looked at two sets.
Birne Binegar: On the Co domain side.
Birne Binegar: we've looked at the range of tea.
Birne Binegar: And it wasn't all the vectors in in the co domain that are that you reach I applying your function there's some vector inside the domain.
Birne Binegar: Let me sort of put a.
Birne Binegar: subset of that, but actually what it's going to be it's going to be a subspace so I shouldn't really be drawing this as a sort of a blob it's going to be.
Birne Binegar: subspace so maybe some hyper plane that's embedded inside this space.
Birne Binegar: But it what it is consistent, all the points that we can reach by applying this function and what we've seen is that this thing is the same thing as the call they sent the matrix and that allows us to calculate this explicitly.
Birne Binegar: there's another subspace but it occurs over here on the domain side.
Birne Binegar: it's the set.
Birne Binegar: Of vectors it's going to be a substations can be some hyper plane in here that contains the origin it consists of all the factors.
Birne Binegar: That go exactly to the zero vector over here that's also a subspace and it's called the kernel of your transformation it works out to be exactly the same thing as the no spaces.
Birne Binegar: At the associated matrix this is going to be a numerical matrix so something we can calculate with will find a basis for the subspace by calculating.
Birne Binegar: The solution set at of X is equal to zero, so now we've got the picture is now complete, not only do I haven't know how to deal with linear transformation basically if I want to calculate anything i'll convert it into a matrix.
Birne Binegar: But then I can relate sub spaces that are defined by their function properties, I can directly calculate them in terms of a numerical matrix.
Birne Binegar: Okay, so let's see.
Birne Binegar: yeah so what i'll stop here today because i'm out of time.
Birne Binegar: What i'm going to do on Friday is i'm going to count really quickly, how many basis vectors I haven't each of these sub faces and then i'll be able to relate the dimensions of the two sub spaces.
Birne Binegar: Okay, so i'll stop here today and we'll finish this up on Friday and then, and then we have spring break and.
Birne Binegar: Probably on Friday will also set the date for the second exam so look at your schedule for the week after spring break and we'll decide one be the best date for the damn they've done that.
yep.
Birne Binegar: yep.